Chapter 3.2, Problem 3E

To determine

To prove: That the function $f$ is not differentiable at the point $P$ . Also, find all points where $f$ is not differentiable.

Answer to Problem 3E

The required point where $f$ is not differentiable at $P\left(1,1\right)$ .

Explanation of Solution

Given Information:

The graph of the function $f$ :

  

Formula Used:

Differentiable function formula:

The differentiable function formulas to find the left hand derivative and right hand derivative are

Left hand derivative $=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ and Right hand derivative $=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ respectively.

Calculation:

From the given graph it is seen that on either side $x=1$ the equation doesn't have the same slope.

Consider both slopes are sided functional derivatives.

So, the limits are not equal.

To find the Left hand derivative, use the differentiable function formula. $\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $\sqrt{x}$ for $f\left(x\right)$ and $1$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 1^{-}}{\lim }\frac{\sqrt{x}-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{-}}{\lim }\frac{\sqrt{x}-1}{x-1}\\ =\underset{x\to 1^{-}}{\lim }\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\\ =\underset{x\to 1^{-}}{\lim }\frac{{\left(\sqrt{x}\right)}^2-1^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\end{array}$

Simplify the above limit further.

  $\begin{array}{c}\underset{x\to 1^{-}}{\lim }\frac{{\left(\sqrt{x}\right)}^2-1^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}=\underset{x\to 1^{-}}{\lim }\frac{\left(x-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\\ =\underset{x\to 1^{-}}{\lim }\frac{1}{\left(\sqrt{x}+1\right)}\\ =\underset{x\to 1^{-}}{\lim }\frac{1}{\left(\sqrt{1}+1\right)}\\ =\frac{1}{2}\end{array}$

Again, to find the Right hand derivative, use the differentiable function formula. $\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $2x-1$ for $f\left(x\right)$ and $1$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 1^{+}}{\lim }\frac{\left(2x-1\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{2x-1-1}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{2x-2}{x-1}\\ =\underset{x\to 1^{+}}{\lim }\frac{2\left(x-1\right)}{x-1}\end{array}$

Simplify the above limit further.

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }\frac{2\left(x-1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }2\\ =2\end{array}$

Since, the limits are not equal;

So, the function is not differentiable at the point $P\left(1,1\right)$ .

Hence, the required point where $f$ is not differentiable is $P\left(1,1\right)$ .